Nesting list with a value repeated `n` amount of times

Question

Given the input of n and value. The code is supposed to nest the single element list with value repeated n times. The final output should be a multilevel nested list with single repeated values in all sublists.

This is code-golf, so the shortest code in bytes wins!

Test cases:

n = 1, value = 'a': ['a']
n = 2, value = 'a': ['a', ['a']]
n = 3, value = 'a': ['a', ['a', ['a']]]
n = 4, value = 'a': ['a', ['a', ['a', ['a']]]]
n = 5, value = 'a': ['a', ['a', ['a', ['a', ['a']]]]]
n = 6, value = 'a': ['a', ['a', ['a', ['a', ['a', ['a']]]]]]

• att: Is n always greater than zero? If so, can we take n 0-indexed?
• theorist: Do we need to include the commas in the output? And can the value be input as a string?
• U13-Forward: @theorist Just a sequence is enough, in any way
• U13-Forward: @att Yes. Always bigger than 0
• U13-Forward: @ZaMoC Sure....

Answer 1

Python 2, (削除) 33 (削除ここまで) 30 bytes

-3 bytes thanks to @xnor

lambda n,v:eval('[v,'*n+']'*n)

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• U13-Forward: Wow! Good thought!!
• xnor: It looks like you can just do lambda n,v:eval('[v,'*n+']'*n)
• dingledooper: @xnor Haha! I honestly have no idea how I didn't see that...
• U13-Forward: Even cooler now!

Answer 2

05AB1E, 3 bytes

F‚Ù

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Explanation:

F # Loop the first input amount of times:
 ‚ # Pair the (implicit) second input-value with the current list
 Ù # And uniquify it (which only does something in the first iteration,
 # transforming the pair of values to a single wrapped value)
 # (after the loop, the result is output implicitly)

• U13-Forward: 3 bytes! So far I think you're the winner.

Answer 3

Vyxal r, 4 bytes

(0"U

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The joys of golfing language :)

Explained

(0"U
( # n times:
 0 # push the value
 "U # pair and uniquify - a port of the 05ab1e answer

• U13-Forward: Wow... 4 bytes!!
• 12944qwerty: That's so short... how-

Answer 4

Python, 34 bytes

f=lambda x,n:n and[x,f(x,n-1)][:n]

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Thanks to loopy walt for this one.

---

Old answers:

Python, 36 bytes

f=lambda x,n:[x][n>1:]or[x,f(x,n-1)]

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Python, ^{(削除) 36 (削除ここまで)} 35 bytes

f=lambda x,n:n*[1]and[[x]+f(x,n-1)]

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Outputs as a singleton list, but that feels like cheating here.

Python, 39 bytes

f=lambda x,n:[x]+(~-n*[1]and[f(x,n-1)])

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Python, 37 bytes

f=lambda x,n:x+(~-n*[1]and[f(x,n-1)])

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Inputs as a singleton list.

• U13-Forward: Wow! Very clever!
• loopy walt: You can go zero-based and save two bytes in the topmost lambda (dropping >1).
• loopy walt: Well, they don't say no, do they? How about this f=lambda x,n:[x,n<2or f(x,n-1)][:n], then (35 bytes)? ato.pxeger.com/…
• loopy walt: 34 if you are ok with requiring n>0. ato.pxeger.com/… And I'm not dingledooper :-)
• pxeger: @loopywalt sorry, your avatars are too similar ;)

Answer 5

Proton, 25 bytes

n=>v=>((a=>[v,a])*n)([v])

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n=>v=>((a=>[v,a])*n)([v]) This language is stupid
n=> Given n
 v=> and v
 (a=>[v,a]) pair v with the current accumulator
 *n n times
 ( )([v]) and call that on [v]

• U13-Forward: Thanks, very cool. "this language" is stupid lol
• hyperneutrino: @U12-F̉͋̅̾̇orward yep - I made this language, so I know better than anyone just how stupid it is :P
• U13-Forward: Wow! Really! Very interesting. Cool.

Answer 6

Wolfram Mathematica, (削除) 30 (削除ここまで) 21 bytes

Print[""@@#~Nest~##]&

-9 bytes from @att!

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Sample I/O:

Print[""@@#~Nest~##]&@@{a,3}

[a[a[a]]]

Print[""@@#~Nest~##]&@@{17,3}

[17[17[17]]]

Print[""@@#~Nest~##]&@@{Pi,6}

\$[\pi [\pi [\pi [\pi [\pi [\pi ]]]]]]\$

• att: Is a separator not needed? 21 bytes if so, or 27 bytes if it is.
• att: (with separator only works if value is a string)

Answer 7

R, 46 bytes

function(n,v)Reduce(list,rep(v,n-1),list(v),T)

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Non-recursive approach. Pretty-printing and test harness taken from Dominic van Essen who insisted I post this as my own.

• Dominic van Essen: Nice; and based on the OP's reply to ZaMoC's comment/question, you can probably leave-out the ,T to save 2 bytes.

Answer 8

BQN, 9 bytes

∾⟜⋈ ́⥊⟜⋈⟜⋈

^{Pins and bowties FTW}

Anonymous tacit function that takes two arguments and returns a nested list. This is why the list formatting looks weird. Run it online!

Explanation

The left argument is the count; the right argument is the value. The example uses a left argument of 2 and a right argument of 0.

∾⟜⋈ ́⥊⟜⋈⟜⋈
 ⟜⋈ Wrap the right argument in a list: ⟨ 0 ⟩, and then
 ⟜⋈ Wrap that list in a list: ⟨ ⟨ 0 ⟩ ⟩, and then
 ⥊ Reshape to a length equaling the left argument: ⟨ ⟨ 0 ⟩ ⟨ 0 ⟩ ⟩
 ́ Right-fold that list on this function:
 ⟜⋈ Wrap the right argument in a list: ⟨ ⟨ 0 ⟩ ⟩, and then
∾ Concatenate with the left argument: ⟨ 0 ⟩ ∾ ⟨ ⟨ 0 ⟩ ⟩ → ⟨ 0 ⟨ 0 ⟩ ⟩

Answer 9

Husk, 13 bytes

+J',R1e'[2R']

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Not an ideal challenge for a language that doesn't support ragged lists...

+ # join together:
 R1 # arg1 repeats of
 e # 2-element list of
 '[ # '[' and
 2 # arg2,
 J', # joined by commas,
 # and
 R'] # arg1 repeats of ']'

Answer 10

Jelly, 5 bytes

W{Ɱṭ/

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Value on left, \$n\$ on right

W The value wrapped in a singleton list
 {Ɱ for each 1..n.
 ṭ/ Reduce by Funky Reverse AppendTM.

• U13-Forward: Just curious, how do you type those characters? :)
• Unrelated String: @U12-F̉͋̅̾̇orward abrudz.github.io/lb/jelly
• U13-Forward: Wow!.... Interesting

Answer 11

Ruby, (削除) 29 (削除ここまで) 23 bytes

->n,a{eval'[a,'*n+?]*n}

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Stole the eval trick from dingledooper

Answer 12

R, 49 bytes

f=function(n,v)`if`(n-1,list(v,f(n-1,v)),list(v))

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• Giuseppe: 46 bytes
• Dominic van Essen: @Giuseppe - that's a different answer. Post it.
• Giuseppe: If you insist!

Answer 13

Racket, 49 bytes

(define(f x n)(if(= n 1)`(,x)`(,x,(f x(- n 1)))))

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• rydwolf: Welcome to Code Golf! Nice first answer.

Answer 14

Wolfram Language (Mathematica), 24 bytes

t&@Do[t={#,t},Set@t;#2]&

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Input [n, value].

Answer 15

Wolfram Language (Mathematica), 24 bytes

0-indexed

Fold[List,{#},Table@##]&

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• att: The order is reversed.
• ZaMoC: @att I guess we will wait for OP to decide...
• hyperneutrino: The challenge specification clearly shows which order it should be in, which was OP's decision when they made the challenge in the first place.
• ZaMoC: @hyper-neutrino OP says my answer is valid. Please check the main comments

Answer 16

MathGolf, 5 bytes

a*Åαç

Outputs reversed (e.g. [[["a"],"a"],"a"] instead of ["a",["a",["a"]]]), which is allowed based on the comments under the challenge.

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Explanation:

a # Wrap the (implicit) input-string into a list
 # e.g. "abc" → ["abc"]
 * # Repeat it the (implicit) input-integer amount of times
 # e.g. 5 → ["abc","abc","abc","abc","abc"]
 Å # For-each over these strings,
 # using the following 2 characters as inner code-block:
 α # Wrap the top two values into a list
 # (which will wrap with the implicit loop-index 0 in the first iteration)
 # e.g. [0,"abc"] in the first iteration
 # [["abc"],"abc"] in the second iteration
 # [[["abc"],"abc"],"abc"] in the third, etc.
 ç # Remove all 0s from the list with a falsey filter
 # (only relevant for the first iteration: [0,"abc"] → ["abc"])
 # (after the loop, the entire stack is output implicitly as result)

Answer 17

HBL, 11 bytes

1,(?(-.)(1('?(-.),)?)?

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Explanation

A recursive function:

1,(?(-.)(1('?(-.),)?)?
1 Cons
 , the second argument with:
 (?(-.) If the first argument decremented is truthy (> 1):
 (1 ) Cons
 ('? ) Recursive call with
 (-.) First argument decremented
 , Second argument unchanged
 ? with nil (empty list)
 Else (the first argument is 1):
 ? Nil (empty list)

I.e., if the first argument is 1, we get (cons arg2 nil), which simply creates a singleton list containing the second argument; and if the first argument is greater than 1, we get (cons arg2 (cons [recursive-call] nil)), which wraps the result of the recursive call in a singleton list and then prepends the second argument to it.

Answer 18

APL+WIN, 28 bytes

Prompts for n followed by value

( ̄1↓∊n⍴⊂'[',⎕,','),(n←⎕)⍴']'

Try it online! Thanks to Dyalog Classic

• Dominic van Essen: From the image it looks as if the outermost list (=box?) doesn't contain an 'a' alement...
• Graham: @DominicvanEssen Thanks for picking that up. I could not find an easy fix so went with a new approach
• Dominic van Essen: Hmm... don't you now have an unwanted comma in the innermost list...?
• Graham: @DominicvanEssen Thanks again. Fixed. Old age seems to be getting the better of me ;(

Answer 19

PHP, (削除) 56 (削除ここまで) 55 bytes

function($n,$a){for($r=[$a];--$n;)$r=[$a,$r];return$r;}

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Like often, PHP makes the worst score, akin only to C, but dirtier and with lots of $

EDIT: -1 byte, these dollars allow us some trickery with the parser after all

Answer 20

PHP 50 bytes

function f($c,$n){return --$n?[$c,f($c,$n)]:[$c];}

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Answer 21

PowerShell Core, 41 bytes

$n,$a=$args
1..$n|%{$r=$a,@($r)|?{$_}}
$r

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Please note that this does not work for n = 1, for some reason PowerShell treats it as a string. Even when forcing to return an array.

Let me know if not OK and I'll withdraw this answer!

• U13-Forward: Thanks for the answer, it's ok! doesn't matter much
• U13-Forward: Or you could add an extra code for an if else for whether n is 1. Then just return the value in a singular list
• Julian: I tried but PowerShell doesn't want it: With an explicit .Net list for example

Answer 22

Pari/GP, 29 bytes

f(a,n)=if(n--,[a,f(a,n)],[a])

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Answer 23

Charcoal, 15 bytes

FN≔⟦+⟦η⟧υ⟧υ⭆1⊟υ

Try it online! Link is to verbose version of code. Explanation:

FN

Repeat n times...

≔⟦+⟦η⟧υ⟧υ

... prepend value to the initially predefined empty list, then wrap that in another list.

⭆1⊟υ

Unwrap the very last wrapper list and pretty-print it.

Answer 24

Pip -p, (削除) 14 (削除ここまで) 13 bytes

-1 byte by porting Neil's Charcoal answer

Lal:[lPEb]l@0

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Explanation

Lal:[lPEb]l@0
 l is empty list; a,b are command-line args
La Loop a times:
 b b
 PE Prepended to
 l Current list
 [ ] Wrap that result in a singleton list
 l: Assign back to l
 l After the loop, l is our desired result wrapped in a singleton list
 @0 So get the first element
 and autoprint it, formatted as a list (-p flag)

Answer 25

JavaScript (ES6), 24 bytes

-2 thanks to @tsh

n=>g=x=>--x?[n,g(x)]:[n]

• tsh: Why not --x instead.

Answer 26

JavaScript (ES6), 27 bytes

f=(c,n)=>--n?[c,f(c,n)]:[c]

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Answer 27

Lua, 50 bytes

n,v=...t={v}for i=2,n do t={v,t}n=n-1 end return t

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Answer 28

Retina, 31 bytes

L$`¶
$`*$([$', )$`*]
, (]+)$
1ドル

Try it online! Takes n and value on separate lines. Explanation:

L$`¶

Match the newline between n and value. This puts n in $` and value in $'.

$`*$([$', )$`*]

Wrap value in n lists.

, (]+)$
1ドル

Remove the trailing comma in the innermost list.

Answer 29

C (clang), 56 bytes

i;f(*x,n){for(i=~n;++i<n;)printf(i<0?",[%s"-i/n:"]",x);}

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Takes a string literal as value and n , prints to std out the result.

for(i=~n;++i<n;) - iterate from -n to n
printf(i<0? - select format string:
",[%s"-i/n..x) * new nest, -i/n to skip comma at first stage
:"]" * close nest
```

Answer 30

Python, 38 bytes:

f=lambda n,v:[v,f(n-1,v)]if n>1else[v]

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• pxeger: This doesn't work for n=1
• U13-Forward: @pxeger Edited mine!
• U13-Forward: 36 bytes now...
• pxeger: You can't remove the f= because this is a recursive function which requires itself to be assigned to f
• U13-Forward: @pxeger dingledooper did